What is the value of the following logarithm? $\log_{15} \left(\dfrac{1}{225}\right)$
Answer: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $15^{y} = \dfrac{1}{225}$ In this case, $15^{-2} = \dfrac{1}{225}$, so $\log_{15} \left(\dfrac{1}{225}\right) = -2$.